Other Points - CLEP Calculus

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Question

Find the critical points (rounded to two decimal places):

Answer

To find the critical points, set and solve for .

Differentiate:

$f{}'(x)= 3x^2+6x+1$

Set equal to zero:

Solve for using the quadratic formula: $x= \frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$a=3,\ b=6,\ c=1$

$x= \frac{-6\pm \sqrt{(6)^2-4(3)(1)}}{2(3)}$

$x= \frac{-6\pm \sqrt{36-12}}{6}}$

$x= \frac{-6+\sqrt{24}}{6}}\ and\ x= \frac{-6-\sqrt{24}}{6}}$

$c=-0.18\ and\ c=-1.82$

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Question

Find the value(s) of the critical point(s) of

.

Answer

In order to find the critical points, we must find $\ f'(x)$ and solve for $\ x$

$\ \ f'(x)=3x^2+4x+1$

Set

$\ \ 3x^2+4x+1=0$

Use the quadratic equation to solve for $\ x$ .

Remember that the quadratic equation is as follows.

$\ \ x=\frac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}$ , where a,b and c refer to the coefficents in the

equation .

In this case, , , and .

After plugging in those values, we get

$\ \ x=\frac{-4\pm\sqrt{4^2-4\cdot3\cdot1}}{2\cdot3} =\frac{-4 \pm \sqrt{4}}{6}=\frac{-4 \pm 2}{6}$ .

So the critical points values are:

$\ \ c_1=\frac{-4+2}{6}=\frac{-1}{3}$

$\ \ c_2=\frac{-4-2}{6}=-1$

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Question

Find the value(s) of the critical point(s) of

.

Answer

In order to find the critical points, we must find $\ f'(x)$ and solve for $\ x$ .

$\ \ f(x)=5x^3+10x^2+x+100$

$\ \ f'(x)=15x^2+20x+1$

Set $\ f'(x)=0$

$\ \ 15x^2+20x+1=0$

Use the quadratic equation to solve for $\ x$ .

Remember that the quadratic equation is as follows.

$\ \ x=\frac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}$ , where a,b and c refer to the coefficents in the equation .

In this case, , , and .

After plugging in those values, we get.

${}\ x=\frac{-20\pm\sqrt{20^2-4\cdot15\cdot1}}{2\cdot15} =\frac{-20 \pm \sqrt{340}}{30} \ \ x=\frac{-20 \pm 2\cdot\sqrt{85}}{30}$

So the critical points values are,

${}\ \ c_1=\frac{-20+2\cdot \sqrt{85}}{30}=\frac{-10+\sqrt{85}}{15} \ \ c_2=\frac{-20-2\cdot \sqrt{85}}{30}=\frac{-10-\sqrt{85}}{15}$

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Question

Find the critical points of

Answer

First we need to find .

Now we set

Now we can use the quadratic equation in order to find the critical points.

Remember that the quadratic equation is

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ ,

where a,b,c refer to the coefficients in the equation

In this case, a=3, b=6, and c=1.

$x=\frac{-6\pm\sqrt{(-6)^2-4(3)(1)}}{(2)(3)}$

$x=\frac{-6\pm\sqrt{24}}{6}=\frac{-6\pm2\sqrt{6}}{6}$

Thus are critical points are

$c_1=\frac{-6+2\sqrt{6}}{6}=-1+\frac{\sqrt{6}}{3}$

$c_2=\frac{-6-2\sqrt{6}}{6}=-1-\frac{\sqrt{6}}{3}$

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Question

Find the critical points of

.

Answer

In order to find the critical points, we need to find using the power rule $f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ .

Now we set , and solve for .

Thus is a critical point.

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Question

Find the critical point(s) of $f(x)=2x^{2}-5x+4$ .

Answer

To find the critical point(s) of a function , take its derivative , set it equal to , and solve for .

Given $f(x)=2x^{2}-5x+4$ , use the power rule

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ to find the derivative. Thus the derivative is, .

Since :

$x=\frac{5}{4}$

The critical point is $c=\frac{5}{4}$ .

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Question

Find the critical points of

.

Answer

In order to find the critical points, we must find using the power rule $f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ .

.

Now we set .

Now we use the quadratic equation in order to solve for .

Remember that the quadratic equation is as follows.

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ ,

where a,b,c correspond to the coefficients in the equation

.

In this case, a=9, b=-40, c=4.

$x=\frac{40\pm\sqrt{(-40)^2-4(9)(4)}}{2(9)}$

$x=\frac{40\pm\sqrt{1600-144}}{18}$

$x=\frac{40\pm\sqrt{1456}}{18}=\frac{40\pm4\sqrt{91}}{18}$

Then are critical points are:

$c_1=\frac{40+4\sqrt{91}}{18}=\frac{20+2\sqrt{91}}{9}$

$c_2=\frac{40-4\sqrt{91}}{18}=\frac{20-2\sqrt{91}}{9}$

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Question

Find all the critical points of

.

Answer

In order to find the critical points, we first need to find using the power rule $f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ ..

Now we set .

$x^2=\frac{2}{3}$

$x=\pm\sqrt{\frac{2}{3}}$

Thus the critical points are at

$x=\sqrt{\frac{2}{3}}$ , and

$x=-\sqrt{\frac{2}{3}}$ .

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Question

Find the critical points of the following function:

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Answer

To find critical points the derivative of the function must be found.

$y=\frac{x^6}{6}+\frac{3x^5}{5}-\frac{13x^4}{4}-5x^3$

Critical points occur where the derivative equals zero.

$x=\left { {{-5, -1, 0, 3}} \right }$

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Question

Determine the point on the graph that is not changing if $f(x)=x^{2}-12x+32$ .

Answer

To find the point where the graph of is not changing, we must set the first derivative equal to zero and solve for .

To evaluate this derivate, we need the following formulae:

$\frac{d}{dx}(c)=0, \frac{d}{dx}(x)=1, \frac{d}{dx}cx=c$

$\frac{d}{dx}(x^{n}))=nx^{n-1}, \frac{d}{dx}(cx^{n})=ncx^{n-1}$

$f'(x)=2x^{2-1}-12x^{1-1}+0=2x-12$

Now, setting the derivate equal to to find where the graph is not changing:

Now, to find the corresponding value, we plug this value back into :

$f(6)=6^{2}-12(6)+32=36-72+32=-4$

Therefore, the point where is not changing is

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Question

Find the limit: $\lim_{x\rightarrow 0} \frac{\sin(6x)}{x}$

Answer

To evaluate this limit, we must use L'Hopital's Rule:

If $\lim_{x\rightarrow c} \frac{f(x)}{g(x)}=\frac{0}{0}$ , take the derivative of both and and then plug in to obtain $\frac{f'(c)}{g'(c)}$

We will also need the power rule, the derivative of the trigonometric function sine, and the chain rule.

Since when we plug in in the numerator and denominator, we obtain a result of $\frac{0}{0}$ , we can use L'Hopitals rule.

To take the derivative of the numerator we need the chain rule, the derivative of the trigonometric function sine, and the power rule.

$\frac{d}{dx}(ax^{n})=anx^{n-1}, \frac{d}{dx} f(g(x))=f'(g(x))g'(x), \frac{d}{dx} \sin(x)= \cos(x)$

Applying the chain rule to the numerator with $f(x)=\sin(6x)$ and , we see that:

$f'(x)=\cos(6x)$ and $g'(x)=6x^{1-1}=6$ .

Now plugging these into the chain rule, we obtain:

$6\cos(6x)$

Now, to find the derivative of the denominator, we need the power rule again:

$\frac{d}{dx}x=x^{1-1}=1$

Now that we have found the derivative of the numerator and denominator, we can apply L'Hopital's Rule:

$\lim_{x\rightarrow 0} \frac{6\cos(6x)}{1}=6\cos(0)=6$

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Question

Find

$\lim_{x\rightarrow 3} \frac{x^{2}+2x-15}{x-3}$ .

Answer

To evaluate this limit, all we need to do is factor the numerator and then cancel out the factor that is in common using the following formula:

$x^{2}+bx+c=(x+d)(x+e)$

With a simple algebra trick, we will be able to easily plug in for :

$\lim_{x\rightarrow 3} \frac{x^{2}+2x-15}{x-3}=\lim_{x\rightarrow 3}\frac{(x+5)(x-3)}{x-3}$

$=\lim_{x\rightarrow 3} x+5=3+5=8$

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Question

Evaluate the limit:

$\lim_{x\rightarrow\frac{\pi}{4}}\frac{\cos(x)-\sin(x)}{2\cos(2x)}$

Answer

To begin, we need L'Hopital's Rule for this problem which states that if you get $\frac{0}{0}$ when you plug in the value into your function when evalutating the limit, you should take the derivative of both the numerator and the denominator and then try plugging in your value again.

$\lim_{x\rightarrow\frac{\pi}{4}}\frac{\cos(x)-\sin(x)}{2\cos(2x)}=\frac{\cos(\frac{\pi}{4})-\sin(\frac{\pi}{4})}{2\cos(2\frac{\pi}{4})}=\frac{0}{0}$

Since this is the case, we will take the derivative of the numerator and denominator.

To take the derivative of the numerator, we need the differentiation formulas for the trigonometric functions cosine and sine.

$\frac{d}{dx} \sin(x)= \cos(x), \frac{d}{dx} \cos(x)= -\sin(x)$

So, the derivative of the numerator is $-\sin(x)-\cos(x)$

To find the derivative of the denominator, we again need the differentiation formula for cosine, as well as the chain rule.

$\frac{d}{dx} f(g(x))=f'(g(x))g'(x)$

In this problem, $f(x)=\cos(2x)$ and

$\frac{d}{dx}\cos(2x)=-\sin(2x)$

$\frac{d}{dx}2x=2$

So, plugging these into the chain rule, we obtain:

$-2\sin(2x)$

Now let's put these expressions back into the numerator and denominator and again try to plug in our limit value:

$\lim_{x\rightarrow \frac{\pi}{4}}\frac{-\sin(x)-\cos(x)}{-4\sin(2x)}$

$=\frac{-\sin(\frac{\pi}{4})-\cos(\frac{\pi}{4})}{-4\sin(2\frac{\pi}{4})}=\frac{\frac{\sqrt2}{2}+\frac{\sqrt2}{2}}{4}=\frac{\sqrt2}{4}$

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Question

Find all points on the graph of $f(x)=\cos^{2}(x)$ where the tangent line is horizontal.

Answer

To solve this problem, we need the chain rule, the derivative of the trigonometric function cosine, and the power rule.

First let's apply the chain rule, which states:

$\frac{d}{dx} h(i(x))=h'(i(x))i'(x)$

In this problem, $h(x)=\cos^{2}(x)$ and $i(x)=\cos(x)$ .

To find , we need the power rule which states:

$\frac{d}{dx}(x^{n})=nx^{n-1}$

$h'(x)=2\cos(x)^{2-1}=2\cos(x)$

To find , we need the derivative of cosine which states:

$\frac{d}{dx} \cos(x)= -\sin(x)$

$i'(x)=-\sin(x)$

Plugging these equations into the chain rule we obtain:

$f'(x)=2\cos(x)-\sin(x)$

To find all points where the tangent line is horizontal, we must first take the derivative of the function and then set it equal to zero:

Setting this equal to zero, we obtain:

$0=2\cos(x)-\sin(x)$

$0=\cos(x)\sin(x)$

Therefore, either $\cos(x)=0$ or $\sin(x)=0$

Recall that from the unit circle, cosine equals zero at $\frac{\pi}{2}, \frac{3\pi}{2}$ and sine equals zero at $0, \pi, 2\pi$ .

So, at every multiple of $\frac{\pi}{2}$ , either $\cos(x)=0$ or $\sin(x)=0$ .

Therefore, $x=\frac{n\pi}{2}$ because at each multiple of $\frac{\pi}{2}$ , either $\cos(x)=0$ or $\sin(x)=0$

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Question

Evaluate the limit:

$\lim_{x\rightarrow2}\frac{\ln(x-1)}{x-2}$

Answer

To solve this problem, we need L'Hopital's Rule, the derivative of the natural logarithm, the chain rule, the power rule, and the derivative of a constant.

Notice that if we plug in our value into the function, we obtain a value of $\frac{0}{0}$ .

$\frac{\ln(2-1)}{2-2}=\frac{\ln(1)}{0}=\frac{0}{0}$

L'Hopital's Rule, which states that if you plug in your limit value and obtain $\frac{0}{0}$ , you should take the derivative of the numerator and denominator and try plugging in your limit value again.

So we will take the derivative of the numerator and denominator.

For the numerator, we need the chain rule,the derivative of the natural logarithm, the derivative of a constant, and the power rule, which state:

$\frac{d}{dx}(c)=0, \frac{d}{dx} f(g(x))=f'(g(x))g'(x), \frac{d}{dx} \ln(x)=\frac{1}{x}$

For the numerator, $f(x)=\ln(x-1)$ and .

Applying the chain rule to this expression yields:

$\frac{d}{dx} \ln(x-1)=\frac{1}{x-1}*(1x^{1-1}-0)=\frac{1}{x-1}$

To find the derivative of the denominator, we need the power rule and the derivative of a constant.

$\frac{d}{dx} x-2=1x^{1-1}-0=1$

So now we have:

$\lim_{x\rightarrow2}\frac{\frac{1}{x-1}}{1}=\lim_{x->2}\frac{1}{x-1}=\frac{1}{2-1}=\frac{1}{1}=1$

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Question

Find the x-coordinate of the critical points of .

Answer

We need to differentiate term by term, applying the power rule,

$(x^n)'=nx^{n-1}$

This gives us

$\begin{align} (x^3-15x^2+75x-45)' &= 3x^2-30x+75 \end{align}$

The x-coordinate of the critical points are the points where the derivative equals 0. To find those, we can use the quadratic formula:

$\begin{align} x &= \frac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}\ &=\frac{30\pm\sqrt{900-4\cdot 3\cdot 75}}{2\cdot 3} \ &=\frac{30\pm\sqrt0}{6}\ &=5 \end{align}$

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Question

Find the x value of the critical points of .

Answer

We need to differentiate term by term, applying the power rule,

$(x^n)'=nx^{n-1}$

This gives us

$\begin{align} (x^3-3x^2-2)' &= 3x^2-6x \end{align}$

The critical points are the points where the derivative equals 0. To find those x values, we can use the quadratic formula:

$\begin{align} x &= \frac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a} \ &= \frac{6\pm\sqrt{36-4\cdot 3\cdot 0}}{6}\ &=\frac{0}{6} \text{ or } \frac{12}{6} \ &=0 \text{ or } 2 \end{align}$

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Question

Find the x values of the critical points of .

Answer

We need to differentiate term by term, applying the power rule,

$(x^n)'=nx^{n-1}$

This gives us

$\begin{align} (2x^3-6x+2)' &= 6x^2-6 \end{align}$

The critical points are the points where the derivative equals 0. To find those, we can use the quadratic formula:

$\begin{align} x &= \frac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}\ &=\frac{0\pm\sqrt{0-4\cdot 6\cdot -6}}{2\cdot 6} \ &=\frac{0\pm\sqrt{144}}{12}\ &=\pm1 \end{align}$

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Question

Find the x values for critical points of .

Answer

We need to differentiate term by term, applying the power rule,

$(x^n)'=nx^{n-1}$

This gives us

$\begin{align} (2x^3+3x^2-12x+5)' &= 6x^2+6x-12 \end{align}$

The critical points are the points where the derivative equals 0. To find those, we can use the quadratic formula:

$\begin{align} x &= \frac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a} \ &= \frac{-6\pm\sqrt{36-4\cdot 6\cdot -12}}{12}\ &= \frac{-6\pm\sqrt{384}}{12}\ &= \frac{-6\pm18}{12}\ &=\frac{12}{12} \text{ or } \frac{-24}{12} \ &=1 \text{ or } -2 \end{align}$

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Question

Find the critical points (rounded to two decimal places):

Answer

To find the critical points, set and solve for .

Differentiate:

$f{}'(x)= 3x^2+6x+1$

Set equal to zero:

Solve for using the quadratic formula: $x= \frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$a=3,\ b=6,\ c=1$

$x= \frac{-6\pm \sqrt{(6)^2-4(3)(1)}}{2(3)}$

$x= \frac{-6\pm \sqrt{36-12}}{6}}$

$x= \frac{-6+\sqrt{24}}{6}}\ and\ x= \frac{-6-\sqrt{24}}{6}}$

$c=-0.18\ and\ c=-1.82$

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